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Reliable solution of an unilateral contact problem with friction and uncertain data in thermo-elasticity
Mathematics and Computers in Simulation 67 (2005) 559–580
Reliable solution of an unilateral contact problem with friction and uncertain data in thermo-elasticity I. Hlav´acˇ ek, J. Nedoma∗ Institute of Computer Science AS CR, Pod Vod´arenskou vˇezˇi´ 2, 182 07 Prague 8, Czech Republic Received 7 July 2004; accepted 2 August 2004 Available online 14 October 2004
Abstract An unilateral contact problem with friction and with uncertain input data in quasi-coupled thermo-elasticity is analysed. As uncertain data coefficients of stress–strain law, coefficients of thermal conductivity, body and surface forces, thermal sources and friction coefficients are assumed, being prescribed in a given set of admissible functions. Method of worst scenario is applied to find the most “dangerous” admissible input data. Stability of the solution with respect to the data is proved and employed for the proof of existence of a solution to the worst scenario problems. Some models in geomechanics, geodynamics and mechanics as well as in technology are stated and the safety of the high level radioactive waste repositories is considered. © 2004 IMACS. Published by Elsevier B.V. All rights reserved. Keywords: Unilateral contact; Steady-state heat flow; Coulomb friction; Finite element analysis; Radioactive waste repositories
1. Introduction In this paper, we will deal with coercive contact problem with friction (see [4,5,7–9]) and uncertain input data in linear quasi-coupled thermo-elasticity, which represents extension of problems solved in [8,9]. Note that a related semi-coercive problem has been treated in a recent paper [4]. By uncertain data we mean input data (physical coefficients, right-hand sides, boundary values, friction limits, etc.), which ∗
Corresponding author. Tel.: +420 266 053 580; fax: +420 286 585 789. E-mail address: [email protected] (J. Nedoma).
0378-4754/$30.00 © 2004 IMACS. Published by Elsevier B.V. All rights reserved. doi:10.1016/j.matcom.2004.08.001
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cannot be determined uniquely but only in some intervals with the errors determined by their measurement errors. The notation reliable solution denotes the worst case among a set of possible solutions, where possibility is given by uncertain input data, and the degree of badness is measured by a criterion-functional (see [3,1]). The main goal of our computation will be to find maximal values of this functional depending on the solution of the problem to be solved. Then we formulate and analyze a corresponding maximization (worst scenario) problem. To its solution we can use methods and algorithms known from optimal design. The solvability of the corresponding worst scenario (maximization) problem is proved. Section 2 is devoted to the formulation of: (i) a stationary heat conduction problem P1 ; and (ii) of the unilateral contact problem with friction P2 , which involves the solution of problem P1 . Sets of admissible uncertain input data are introduced in Section 3 and a unique solvability of problems P1 and P2 is proved for any admissible data. In Section 4, we choose six different criteria, which enable us to find the “worst” data. They represent mean values of: (i) the temperatures or displacements over a priori chosen small subdomains; (ii) the intensity of shear stress; and (iii) the normal component of the stress vector on the contact zone. Section 5 contains the proof of continuous dependence of the temperature and displacement functions on the input data. Existence of at least one solution to each of the worst scenario problems is proved in Section 6.
2. Formulation of the thermo-elastic contact problem Let us consider a union Ω of bounded domains Ωι , ι = 1, . . . , s, with Lipschitz boundaries ∂Ωι , occupied by elastic bodies such that Ω = ∪sι=1 Ωι ⊂ RN , N = {2, 3}. Let the boundary ∂Ω = ∪sι=1 ∂Ωι consist of three disjoint parts Γτ , Γu and Γc , such that ∂Ω = Γ¯ τ ∪ Γ¯ u ∪ Γ¯ c . Assume that (N − 1)-dimensional measures of Γτ , Γu and Γckl are positive, where Γc = ∪k,l Γ kl , Γ kl = ∂Ωk ∩ ∂Ωl , 1 ≤ k, l ≤ s, k = l, and Γ¯ τ , Γ¯ u , Γ¯ c denote the closures in ∂Ω. We will deal with the following quasi-coupled problem of thermo-elasticity, which consists of a pair of boundary value and contact problems to be solved one after the other. 2.1. Problem of stationary heat conduction—problem P1 Let W ι and T1 be given functions. Find the function of temperature T = sι=1 T ι such that ι ∂ ι ∂T κij + W ι = 0 in Ωι , 1 ≤ ι ≤ s, i, j = 1, . . . , N, ∂xi ∂xj κij
∂T ni = 0 on Γu , ∂xj
T = T1 , on Γτ , k l ∂T ∂T k l T =T , κij ni + κij ni = 0 on ∪k,l Γ kl , ∂xj ∂xj
(1) (2) (3)
1 ≤ k, l ≤ s.
(4)
Throughout the paper we use the summation convention, i.e. a repeated index implies summation from 1 to N. Furthermore, nk = (nki ), i = 1, . . . , N, 1 ≤ k ≤ s, denotes the unit normal with respect to
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561
∂Ωk , nk = −nl on Γ kl ; (κijι ) is the matrix of thermal conductivities. Assume that κι are positive definite symmetric matrices, 0 < κ0ι ≤ κijι ζi ζj |ζ|−2
for a.a.
x ∈ Ωι , ζ ∈ RN ,
where κ0ι , are constants independent of x ∈ Ωι . Let κijι ∈ L∞ (Ωι ), W ι ∈ L2 (Ωι ), T1 ∈ sι=1 H 1 (Ωι ), T1k = T1l on ∪k,l Γ kl . Definition 1. We say that a function T is a weak solution of problem P1 , if T − T1 ∈ V1 and b(T, z) = s(z) ∀z ∈ V1 , where b(T, z) =
s ι=1
Ωι
V1 = {z ∈ W1 =
(5)
∂zι dx, ∂xi ∂xj
∂T κijι
ι
sι=1 H 1 (Ωι )
s(z) =
s ι=1
Ωι
W ι zι dx,
| z = 0 on Γτ , zk = zl on ∪k,l Γckl }.
The formulation (5) can be obtained by multiplying Eq. (1) by a test function, integrating by parts over the domain Ωι and using the boundary conditions. 2.2. Problem of unilateral contact problem with friction—problem P2 ι Let the body forces F , the surface tractions P, boundary displacements u0 , elastic coefficients cijkl , ι kl ι coefficients of thermal expansion βij and slip limits gc , the temperature T and the reference temperature T0ι = const. be given. We will deal with the following problem.
Problem 1. (P2 ): Find the displacement field u = (ui ), i = 1, . . . , N, in Ω, such that ∂ τij (uι , T ι ) + Fiι = 0 in Ωι , ∂xj
1 ≤ ι ≤ s,
i = 1, . . . , N,
ι τij (uι , T ι ) = cijkl ekl (uι ) − βijι (T ι − T0ι ) in Ωι , 1 ≤ ι ≤ s,
i = 1, . . . , N,
u = u0 on Γu , i = 1, . . . , N,
τnk ≤ 0,
kl kl |τ kl t | ≤ g on ∪k,l Γ ,
(9)
(ukn − ukn )τnk = 0 on ∪k,l Γ kl ,
1 ≤ k, l ≤ s,
1 ≤ k, l ≤ s,
kl
(12) ukt
ult
=
Here, eij (u) = 1/2((∂ui /∂xj ) + (∂uj /∂xi )), ukn = uki nki , k (uti ), ukti = uki − ukn nki , ult = (ulti ), ulti = uli − uln nki , i = 1, k τnk nki , τnl = τijl nli nkj , τ kl t ≡ τt .
uln
= g =⇒ there exists ϑ ≥ 0 such that
(10) (11)
kl |τ kl =⇒ ukt − ult = 0, t |
|τ kl t |
(7) (8)
τij (u, T )nj = Pi on Γτ , ukn − ukn ≤ 0,
(6)
−
−ϑτ kl t .
(13)
= uli nki = −uli nli (no sum over k or l), ukt . . . , N, τnk = τijk nki nkj , τ kt = (τtik ), τtik = τijk nkj
= −
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ι Assume that cijkl are positive definite symmetric matrices such that ι 0 < c0ι ≤ cijkl ξij ξkl |ξ|−2
x ∈ Ωι , ξ ∈ RN ξij = ξij , 2
for a.a.
ι where c0ι , are constants independent of x ∈ Ωι . Let cijkl ∈ L∞ (Ωι ), Fiι ∈ L2 (Ωι ), Pi ∈ L2 (Γτ ), βijι ∈ ι ∞ ι 1 ι 2 L (Ω ), u0 ∈ [H (Ω )] . To simplify the formulation of stress–strain relations, the entries of any symmetric (N × N) matrix {τij } will be denoted by vector notation {τj }, j = 1, . . . , jN , where jN = N(N + 1)/2, as follows:
τi = τii
for 1 ≤ i ≤ N,
τ3 = τ12
for N = 2,
τ4 = τ23 ,
τ5 = τ31 ,
τ6 = τ12 ,
for N = 3.
Likewise, we replace the symmetric matrices (eij (u)), (βij ) by vectors {ej (u)}, {βj }. Then the stress– strain relation (7) can be rewritten as τi (uι , T ι ) =
jN
Aιij ej (uι ) − βiι (T ι − T0ι ),
1 ≤ i, j ≤ jN , 1 ≤ ι ≤ s,
(14)
j=1
where Aι is a symmetric (jN × jN ) matrix, Aιik ∈ L∞ (Ωι ), ι = 1, . . . , s. Assume that Aιik = 0 for i ≤ N and k > N. It is readily seen that τ : e ≡ τij eij =
N
τi ei + 2
i=1
jN
τi ei .
(15)
i=N+1
Therefore, we can write ι eij ekl cijkl
=
jN
Bijι ei ej ,
i,j=1
where Bι is a symmetric (jN × jN ) matrix such that Bijι = Aιij Bijι
=
2Aιij
Bijι = 0
for 1 ≤ i, j ≤ N,
for N + 1 ≤ i, j ≤ jN .
In what follows, we denote W1 = sι=1 H 1 (Ωι ), sι=1 [H 1 (Ωι )]2 ,
(16)
wW1 =
1/2 wι 21,Ωι
ι≤s
W=
for 1 ≤ i ≤ N, N + 1 ≤ j ≤ jN ,
vW =
ι≤s i≤N
, 1/2
vιi 21,Ωι
.
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563
Assume that the matrices Bι are positive definite, so that 0<
a0ι
≤
jN
Bijι ξi ξj |ξ|−2 ≤ a1ι < +∞
for a.a.
x ∈ Ωι , ξ ∈ RjN ,
i,j=1
where the constants a0ι , a1ι are independent of x ∈ Ωι . Finally, let u0 ∈ W, T0 ∈ W1 , gckl ∈ L∞ (Γ kl ), βjι ∈ L∞ (Ωι ). Let us introduce the space of virtual displacements V = {v ∈ W : v = 0 on Γu } and the set of admissible displacements K = {v ∈ V : vkn − vln ≤ 0 on ∪k,l Γ kl }. Definition 2. We say that the function u is a weak solution of problem (P2 ), if u − u0 ∈ K and a(u, v − u) + jg (v) − jg (u) ≥ S(v − u, T ) where a(u, v) =
s ι=1
jg (v) =
k,l
S(v, T ) =
jN Ωι i,j=1
Γ kl
∀v ∈ u0 + K,
Bijι ei (uι )ej (vι ) dx,
(18)
gkl |vkt − vlt | ds
s ι=1
Ωι
(Fiι vιi
(17)
(19) ι
+ (T −
T0ι )βι
ι
: e(v )) dx +
Γτ
Pi vi ds.
(20)
Remark 1. Note that we have to insert the weak solution T of the problem (P1 ) in S(v, T ). Remark 2. The formulation (17) can be obtained by multiplying Eq. (6) by a test function, integrating by parts over the domain Ωι and using the boundary conditions. In Definition 2 we assume that u0 satisfies conditions uk0n − ul0n = 0 on ∪k,l Γ kl .
(21)
Assumption 1. Throughout the paper, we will assume that measN−1 (Γu ∩ ∂Ωι ) > 0
and
measN−1 (Γτ ∩ ∂Ωι ) > 0 for all ι = 1, . . . , s.
3. Sets of uncertain input data Let us assume that the input data A = {Bι , κι , W ι , T1 , F ι , βι , P, u0 , gkl , ι = 1, . . . , s, ∀k, l}
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are uncertain. Let the only available information about them be that they belong to some sets of admissible data, i.e., ι
B A ∈ Uad ⇔ Bι ∈ Uad , βι
βι ∈ Uad ,
P P ∈ Uad ,
ι
ι
κ κι ∈ Uad ,
W W ι ∈ Uad ,
u0 u0 ∈ Uad ,
gkl ∈ Uad .
T1 T1 ∈ Uad ,
ι
F F ι ∈ Uad ,
gkl
¯ ι exist such that Assume that all the bodies Ωι are piecewise homogeneous, so that partitions of Ω ι ¯ ι = ∪Jj=1 ¯ jι , Ω Ω kl ¯ kl Γ kl = ∪Q q=1 Γq ,
Ωjι ∩ Ωkι = ∅ for j = k, 1 ≤ ι ≤ s ,
(22)
Γqkl ∩ Γpkl = ∅ for q = p, ∀k, l
(23)
and let the data Bι , κι , F ι , W ι , βι be piecewise constant with respect to the partition (22). Let us denote Γu ∩ ∂Ωι = Γuι ,
ι = 1, . . . , s,
(24)
Γτ ∩ ∂Ωι = Γτι ,
ι ≤ s.
(25)
We define the sets of admissible matrices: ι
B ι Uad = {(jN × jN ) symmetric matrices Bι : Bιik (j) ≤ Bik| Ωι
j
= const. ≤
ι ¯ ik B (j),
ι
j≤J,
i, k = 1, . . . , jN },
(26)
¯ ι (j) are given (jN × jN ) symmetric matrices, ι = 1, . . . , s. Assume that positive where Bι (j) and B ι constants cB (j) exist such that ¯ ι (j) − Bι (j)) ≡ cBι (j) for j = 1, . . . , J ι , ι = 1, . . . , s, (27) ¯ ι (j)) − ρ 1 (B λmin 21 (Bι (j) + B 2 where λmin and ρ denotes the minimal eigenvalue and the spectral radius, respectively. ι Remark 3. If the conditions (26) and (27) are fulfilled, the matrices Bι (j) ≡ B|Ω ι are positive definite j
ι
B , ι = 1, . . . , s and any j ≤ J ι . This follows from a result of Rohn (see [11](Theorem for any Bι ∈ Uad 5)).
Likewise, we define the set of admissible matrices ι
κ ι Uad = {(N × N) − symmetric matrices κι : κιik (j) ≤ κik| Ωι
j
= const. ≤
ι κ¯ ik (j),
ι
j ≤ J , i, k ≤ N},
(28)
where κι (j) and κ¯ ι (j) are given (N × N) symmetric matrices, j = 1, . . . , J ι , ι = 1, . . . , s. Assume that positive constants cκι (j) exist such that λmin 21 (κι (j) + κ¯ ι (j)) − ρ 21 (¯κι (j) − κι (j)) ≡ cκι (j) for j ≤ J ι , ι ≤ s , (29) where λmin and ρ denotes the minimal eigenvalue and the spectral radius, respectively. Then the matrices ι κι ι ι κι (j) = κ|Ω ι are positive definite for any κ ∈ U , ι ≤ s, j ≤ J . ad j Next, let us introduce Fι
Uadi = {f ∈ L∞ (Ωι ) : F ιi (j) ≤ f|Ωjι = const. ≤ F¯ iι (j), j ≤ J ι },
(30)
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565
for i ≤ N, ι ≤ s, where F ιi (j) and F¯ iι (j) are given constants; ι
W ¯ ι (j), j ≤ J ι }, = {w ∈ L∞ (Ωι ) : W ι (j) ≤ w|Ωjι = const. ≤ W Uad
(31)
¯ ι (j) are given constants; for ι ≤ s, where W ι (j) and W T1 = {T ∈ L∞ (Γτ ) : T 1 (ι) ≤ T|Γτι = const. ≤ T¯ 1 (ι), ι ≤ s}, Uad
(32)
where T 1 (ι) and T¯ 1 (ι) are given constants; u0i Uad = {u ∈ L∞ (Γu ) : u0i (ι) ≤ u|Γuι = const. ≤ u¯ 0i (ι), ι ≤ s},
(33)
where u0i (ι) and u¯ 0i (ι), i = 1, . . . , N, are given constants. Qιτ ι Let Γτι = ∪q=1 Γ¯ τqι , Γτp ∩ Γτqι = ∅ for p = q. We define Pi Uad = {p ∈ L∞ (Γτ ) : p|Γτqι ∈ C(0),1 (Γ¯ τqι ),
q = 1, . . . , Qιτ , ι = 1, . . . , s, p − pi0 ∞ ≤ C1i , ∂p/∂s∞ ≤ C2i }, where C1i , C2i are given constants and pi0 is a given function such that pi0|Γτqι ∈ C(0),1 (Γ¯ τqι ),
∀q ≤ Qιτ ,
∀ι ≤ s,
∂pi0 /∂s∞ ≤ C2i .
Let ι
β Uadi = {b ∈ L∞ (Ω) : βιi (j) ≤ b|Ωjι = const. ≤ β¯ ι (j), j ≤ J ι },
(34)
for i ≤ jN , ι ≤ s, where βιi (j) and β¯ iι (j) are given constants; gkl
Uad = {g ∈ L∞ (Γ kl ) : g|Γ¯ qkl ∈ C(0),1 (Γ¯ qkl ); 0 ≤ g(s) ≤ g¯ qkl , |dg/ds| ≤ Cqkl a.e. in Γqkl , q ≤ Qkl }, (35) for all pairs kl under consideration, where g¯ qkl and Cqkl are given positive constants. Here, C(0),1 denotes the space of Lipschitz-continuous functions. Finally, we introduce the set of admissible data, as follows: Fι
βι
Pi T1 B κ W Uad = ι≤s Uad × ι≤s Uad × ι≤s,j≤N Uadi × ι≤s Uad × Uad × ι≤s,i≤N Uad × i≤N Uad ι
ι
gkl
u0i × i≤N Uad × k,l Uad .
ι
(36)
T1 Remark 4. To obtain T1 ∈ W1 , we have to extend the boundary values T1 ∈ Uad into the domains k l ι kl Ω properly, i.e. satisfying the conditions T1 = T1 on all Γ . As a consequence at some intersections T1 Γ kl ∩ Γ¯ τ (if any), additional continuity conditions are necessary in the definition of Uad , namely: T1k = T1l , u 0i k l kl if Γ¯ τ ∩ Γ¯ τ = ∅. An analogous remark holds for the data u0i ∈ Uad and Γ ∩ Γ¯ u .
Definition 3. Instead of the bilinear forms and functionals b(T, z), a(u, v), jg (v), s(z), S(v, T ) introduced in Definitions 1 and 2, we will write b(A; T, z), a(A; u, v), jg (A; v), s(A; z), S(A; v, T ) for any A ∈ Uad .
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Lemma 1. There exist positive constants Ci , i = 0, 1, . . . , 6 independent of A ∈ Uad , such that b(A; z, z) ≥ C0 z2W1
∀z ∈ V1 ,
|b(A; z, y)| ≤ C1 zW1 yW1 a(A; v, v) ≥ C2 v2W
∀z, y ∈ W1 ,
∀v ∈ V,
|a(A; v, w)| ≤ C3 vW wW |s(A; z)| ≤ C4 z0,Ω
(37) (38) (39)
∀v, w ∈ W ,
∀z ∈ V1 ,
(40) (41)
|S(A; v, T )| ≤ C5 (v0,Ω + v0,Γτ + T − T0 0,Ω vW ) ∀v ∈ W , uι − vι 0,∂Ωι , ∀u, v ∈ W. |jg (A; u) − jg (A; v)| ≤ C6
(42) (43)
ι≤s
Proof. By Theorem 5 in [11], we have (see (29)) λmin (κι (j)) ≥ cκι (j)
ι
κ ∀κι ∈ Uad ,
ι ≤ s, j ≤ J ι .
As a consequence, we obtain ι b(A; z, z) ≥ min ι cκ (j) |grad zι |2 dx. ι≤s,j≤J
ι≤s
(44)
Ωι
Next, we have |grad zι |2 dx ≥ C1ι zι 21,Ωι
(45)
Ωι
for any restriction zι of z ∈ V1 , making use of Assumption 1. Combining (44) and (45), we arrive at (37). κι The inequality ((38) follows from the definitions of Uad immediately. Arguing as in (44), we may write jN ι a(A; v, v) ≥ min ι cB (j) e2k (vι ) dx. (46) ι≤s,j≤J
ι≤s
Ωι k=1
Using Assumption 1, the Korn’s inequality e(vι ) : e(vι ) dx ≥ C2ι vι 21,Ωι
(47)
Ωι
holds for any restriction vι of v ∈ V . Since jN
1 e2k e:e≤ 2 k=1
(48)
(recall formula (15)), combining (46)–(48), we obtain the inequality (39). The inequality (40) is an easy Bι . consequence of the definitions of Uad
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Likewise, we may write ι ι ¯ |s(A; z)| ≤ maxι {|W (j)|, |W (j)|} |zι | dx ≤ C4 z0,Ω . j≤J
ι≤s
Next, we have |S(A; v, T )| ≤
Ωι
[N 1/2 max{|F¯ iι (j)|, |F ιi (j)|} vι 0,Ωι ι≤s
+ C max β¯ ι (j)
Ωι
ι
2|T −
T0ι | e(vι ) dx
+ C(p0 ) vι 0,Γτι ]
≤ C5 v0,Ω + v0,Γτ + T − T0 0,Ω vW . Finally, we may write |jg (A; u) − jg (A; v)| ≤
Γ kl
k,l
≤
k,l
≤C
gkl |(ukt − vkt ) − (ult − vlt )| ds
max g¯ qkl
q≤Qkl
Γ kl
(|ukt − vkt | + |ult − vlt |) ds
(uki − vki 0,Γ kl + uli − vli 0,Γ kl ) ≤ C6 uι − vι 0,∂Ωι . k,l i≤N
ι≤s
Proposition 1. There exists a unique weak solution T (A) of the problem (P1 ) for any A ∈ Uad . Moreover, T (A)W1 ≤ c,
(49)
where c is independent of A. Proof. We can use the Lax–Milgram Theorem, since (5) can be written as b(t, z) = s(z) − b(T1 , z)
∀z ∈ V1
(50)
for t = T (A) − T1 ∈ V1 . Indeed, the form b is V1 -elliptic due to (37) and continuous due to (38), the right-hand side is a linear continuous functional on V1 , by virtue of (41) and (38). Moreover, inserting z := t tW1 ≤ (C4 + C1 T1 W1 )/C0 follows from (50). Using the inverse theorem on traces (see [6, Theorem 2.5.7]), we obtain ˆ 1ι 1 ,∂Ωι ≤ C ˆC ˜ ι max{|T¯ 1 (ι)|, |T 1 (ι)|} T1ι 1,Ωι ≤ CT 2
for any ι ≤ s. As a consequence, T1 W1 is bounded by a constant independent of A ∈ Uad , so that T (A)W1 ≤ T1 W1 + (C4 + C1 T1 W1 )/C0 ≤ C follows for any A ∈ Uad .
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I. Hlav´acˇ ek, J. Nedoma / Mathematics and Computers in Simulation 67 (2005) 559–580
Proposition 2. There exists a unique weak solution u(A) of problem (P2 ) for any A ∈ Uad . Proof. For brevity, let us omit “A” in what follows. The variational inequality (17) is equivalent with the minimization of the functional Φ(u) = 21 a(u, u) + jg (u) − S(u, T ) over the set u0 + K (cf. [2](Chapter 4, Theorem 0.5)). In fact, u0 + K is a convex set, a(u, u) and S(u, T ) are differentiable with respect to u and jg is convex. By Lemma 1 and the trace theorem, we have |jg (u) − jg (v)| ≤ C6
uι − vι 0,∂Ωι ≤ C
ι≤s
uι − vι 1,Ωι ≤ Cs 2 uι − vι W . 1
ι≤s
As a consequence, the functional Φ is continuous and convex, so that it is weakly lower semi-continuous. The functional is coercive on u0 + K. Indeed, K ⊂ V , so that Lemma 1 yields Φ(u0 + v) = 21 a(u0 + v, u0 + v) − S(u0 + v, T ) ≥ 21 a(v, v) + a(u0 , v) − S(u0 + v, T ) ≥ 21 C2 v2W − C7 vW − C8
∀v ∈ K,
since we may write ˆ |S(u0 + v, T )| ≤ C5 C(u 0 + vW + T − T0 0,Ω u0 + vW ) ˆ ≤ C5 C(u 0 W + vW )(1 + T 0,Ω + T0 0,Ω ) ≤ C7 vW + C8 by virtue of (49). The set u0 + K is weakly closed, being convex and closed in W. As a consequence, there exists a minimizer u(A) of Φ over u0 + K (see [2](Chapter 4, Theorem 0.2)). To verify the uniqueness, assume that u = u0 + v and w = u0 + ω are two solutions of (17). Then we may write w − u = ω − v ∈ V and a(u, w − u) + jg (w) − jg (u) ≥ S(w − u, T ), a(w, u − w) + jg (u) − jg (w) ≥ S(u − w, T ). By addition and Lemma 1, we arrive at C2 w − u2W ≤ a(w − u, w − u) ≤ 0, so that w = u follows.
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4. Criteria of worst scenario To find the “worst”, i.e. the most “dangerous” input data A in the set Uad , we need a criterion, i.e. a functional, which depends on the solution T (A) or u(A) of problem (P1 ) or (P2 ), respectively. Next, we present several examples of such criteria. Let Gr ⊂ ∪ι≤s Ωι , r = 1, . . . , r¯ , be (small) subdomains, adjacent to the boundaries ∂Ωι , for example. We can define Φ1 (T ) = max ϕr (T ),
(51)
r≤¯r
where ϕr (T ) = (measN Gr )−1
Gr
T dx; let Gr ⊂ Γu , r ≤ r¯ and
Φ2 (T ) = max ψr (T ) ,
(52)
r≤¯r
where ψr (T ) = (measN−1 Gr )−1
Gr
T ds. Next, we may define
Φ3 (u) = max χr (u) ,
(53)
r≤¯r
with χr (u) = (measN Gr )−1 Gr ui ni (Xr ) dx, where n(Xr ) is the unit outward normal at a fixed point Xr ∈ ∂Ωι ∩ ∂Gr (if Gr ⊂ Ωι ) to the boundary ∂Ωι ; Φ4 (u) = max χr (u) ;
(54)
r≤¯r
where χr (u) = (measN−1 Gr )−1 G ui ni (Xr ) ds; Gr = ∪ι≤s ∂Ωι \Γu . r Since the weak solution u(A) of our problem (17) depends on T (A), then u(A) = u(A; T (A)) and instead of Φi (u) we write Φi (A; u, T ). Another choice is Φ5 (A; u, T ) = max ωr (A; u, T ) ; r≤¯r
(55)
where ωr (A; u, T ) = (measN Gr )−1 Gr I22 (τ(A; u, T )) dx. Here, I2 (τ) denotes the intensity of shear stress (see, e.g. [7]), i.e. the second fundamental invariant of the stress tensor deviator τ D . We have the formulae I22 (τ) =
3
τijD τijD ,
τijD = τij − 13 τkk δij ;
i,j=1 2 2 2 2 2 2 I22 = 23 [τ11 + τ22 + τ33 − (τ11 τ22 + τ11 τ33 + τ22 τ33 ) + 3(τ12 + τ13 + τ23 )].
In (55), τ(A; u, T ) is defined by the formula (7). For orthotropic material and plane strain, we have to insert τ13 = τ23 = 0. If the friction can be neglected (as in [4,7]), we set g0kl ≡ 0 and define, e.g. Φ6 (A; u, T ) = max µr (A; u, T ); r≤¯r
(56)
where µr (A; u, T ) = (measN Gr )−1 Gr (−τn (A; u, T )) ds, τn = τij ni (Xr )nj (Xr ), Xr ∈ Γc ∩ ∂Gr and Gr is a small subdomain adjacent to Γc (cf. (53)).
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Now we formulate the worst scenario problems: find A0i = arg max Φi (T (A)), A∈Uad
i = 1, 2
(57)
and A0i = arg max Φi (A, u(A), T (A)), A∈Uad
i = 3, . . . , 6,
(58)
where T (A) and u(A) are weak solutions of the problem (P1 ) and (P2 ), respectively.
5. Stability of weak solutions To prove the solvability of worst scenario problems (57) and (58)), we have to study the mapping A → T (A) and A → u(A, T (A)). To these end, we introduce the following decomposition of A ∈ Uad : A = {A , A },
where
A = { ι≤s j≤J ι κι (j), ι≤s j≤J ι W ι (j), ι≤s T1ι },
A ∈ Rp1
p1 = (jN + 1)
J ι + s,
ι≤s
and A = { ι≤s j≤J ι Bι (j), ι≤s j≤J ι F ι (j), ι≤s P ι , ι≤s uι0 , ι≤s j≤J ι βι (j), k,l q≤Qkl gkl (q)}. ι
ι
W κ We are going to show the continuity of the mappings A → T (A ) for A ∈ Uad = ι≤s Uad × ι≤s Uad × T1ι Uad and A → u(A, T (A )) for A ∈ Uad , respectively. , An → A in Rp1 as n → ∞. Then Theorem 1. Let An ∈ Uad
T (An ) → T (A )
in W1 .
Proof. By Proposition 1 the sequence of Tn := T (An ) is bounded. We can choose a weakly convergent subsequence. It is easy to verify that its weak limit coincides with a weak solution T := T (A ). Then the uniqueness implies that Tn D T holds for the whole sequence. To prove its strong convergence, we show that b(A ; Tn − T, Tn − T ) = b(A ; Tn , Tn ) − 2b(A ; Tn , T ) + b(A ; T, T ) → 0
(59)
and make use of the inequality (37). To derive (59), we employ also the “inverse trace theorem” [6](Theorem 2.5.7) in the following form T1 − T1n W1 ≤ CT1 − T1n ∞,Γτ .
(60)
Lemma 2. If An ∈ Uad , An → A in U, where U = Rp1 +p2 × ι≤s q≤Qιτ C(Γ¯ τq ) × k,l q≤Qkl C(Γ¯ qkl ), and un D u weakly in W, then a(An ; un , v) → a(A; u, v)
∀v ∈ W,
(61)
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571
S(An ; un , T ) → S(A; u, T ) ∀T ∈ W1 ,
(62)
jg (An ; un ) → jg (A; u).
(63)
Proof. We have |a(An ; un , v) − a(A; u, v)| ≤ |a(An ; un , v) − a(A; un , v)| + |a(A; un − u, v)| = J1 + J2 ;
jN
ι(n) ι
J1 ≤ (Bik (j) − Bik (j))ei (un )ek (v) dx
ι
ι≤s j≤J ι Ωj i,k=1 ≤ CB(n) − B∞,Ω un W vW → 0, since un are bounded in W; J2 → 0 due to the weak convergence of un . Thus, we are led to (61). Next, we may write |S(An ; un , T ) − S(A; u, T )| ≤ |S(An ; un , T ) − S(A; un , T )| + |S(An ; un − u, T )| = I1 + I2 ;
ι(n) ι(n) ι ι ι ι ι ι(n) ι I1 ≤
(Fi (j) − Fi (j))uni + (T − T0 )(β − β ) : e(un )) dx + (Pi −Pi )uni ds
Ωjι
Γτ ι ι≤s,j≤J
≤ C(F (n) − F ∞,Ω + T − T0 W1 β(n) − β∞,Ω + P (n) − P∞,Γτ )un W → 0, I2 → 0 by virtue of the weak convergence of un . As a consequence, (62) follows. We have |jg (An ; un ) − jg (A; u)| ≤ |jg (An ; un ) − jg (A; un )| + |jg (A; un ) − jg (A; u)| = K1 + K2 , K1 ≤
k,l
Γ kl
kl
(g − gkl ) uk − ul ds ≤ C gnkl − gkl ∞,Γ kl un W → 0. n nt nt k,l
Finally, (43) and the compactness of the trace map H 1 (Ωι ) → L2 (∂Ωι ) yield that K2 ≤ C6
ι≤s
uι − uιn 0,∂Ωι → 0.
Thus, we obtain (63). Theorem 2. Then
Let An ∈ Uad , An → A in U ≡ Rp1 +p2 × ι≤s q≤Qιτ C(Γ¯ τq ) × k,l q≤Qkl C(Γ¯ qkl ).
u(An ) → u(A) in W.
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Proof. For brevity, denote un := u(An ), u := u(A), u0n := u0 (An ), u0 := u0 (A), Tn := T (An ), T := T (A), in what follows. We shall insert u := u0 + w(A),
w(A) ∈ K,
un := u0n + wn
wn ∈ K
v := u0 + w or v := u0n + w, w ∈ K into the variational inequality (17). Thus, we obtain a(An ; wn , w − wn ) + jg (An ; u0n + w) − jg (An ; u0n + wn ) ≥ S(An ; w − wn , Tn ) − a(An ; u0n , w − wn ).
(64)
Taking w = 0 and using Lemma 1, we get C0 wn 2W ≤ a(An ; wn ; wn ) + jg (An ; uon + wn ) ≤ S(An ; wn , Tn ) + a(An ; u0n , −wn ) + jg (An ; u0n ) ≤ C5 (wn 0,Ω + wn 0,Γτ + Tn − T0 0,Ω wn W ) uιon 0,∂Ωι . + C3 u0n W wn W + C6
(65)
ι≤s
u0i , and Remark 4, the extensions uιon are By Theorem 1, Tn are bounded in W1 . By definition of Uad 1 ι bounded in H (Ω ). Then
C0 wn 2W ≤ C7 wn W + C8 follows from (65). As a consequence, wn are bounded in W and there exists a subsequence {wk } and a function ω ∈ W such that wk D ω weakly in W, as k → ∞.
(66)
We can show that ω = w(A). First, ω ∈ K since the set K is weakly closed. Second, we have a(Ak ; wk − ω, wk − ω) ≥ 0 so that a(Ak ; wk , wk − ω) ≥ a(Ak ; ω, wk − ω). By Lemma 2, we obtain lim inf a(Ak ; wk , wk − ω) ≥ lim a(Ak ; ω, wk − ω) = 0 . Inserting w := ω into (64), we get a(Ak ; wk , ω − wk ) + jg (Ak ; uok + ω) − jg (Ak ; uok + wk ) ≥ S(Ak ; ω − wk , Tk ) − a(Ak ; u0k , ω − wk )
(67)
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573
and lim sup a(Ak ; wk , wk − ω) ≤ lim sup S(Ak ; wk − ω, Tk ) + lim sup a(Ak ; u0k , ω − wk ) + lim sup[jg (Ak ; uok + ω) − jg (Ak ; uok + wk )].
(68)
For any A ∈ Uad , v ∈ W and T ∈ W1 , we may write S(A; v, T ) = S1 (A; v) + S2 (A; T − T0 , v), where S1 (A; ·) is a linear continuous functional on W and S2 (A; ·, ·) is a bilinear continuous form on L2 (Ω) × W. Then we have S(Ak ; wk − ω, Tk ) = S1 (Ak ; wk − ω) + S2 (Ak ; Tk − T0 , wk − ω) = S1 (A; wk − ω) + [S1 (Ak ; wk − ω) − S1 (A; wk − ω)] + S2 (A; Tk − T0 , wk − ω) + [S2 (Ak ; Tk − T0 , wk − ω) − S2 (A; Tk − T0 , wk − ω)]. By virtue of the weak convergence, S1 (A; wk − ω) → 0;
S2 (A; Tk − T0 , wk − ω) → S2 (A; T − T0 , 0) = 0
since Tk → T in W1 by virtue of Theorem 1. Next, we have |S1 (Ak ; wk − ω) − S1 (A; wk − ω)| ≤ CAk − AU wk − ωW → 0, |S2 (Ak ; Tk − T0 , wk − ω) − S2 (A; Tk − T0 , wk − ω)| ≤ CAk − AU Tk − T 0,Ω wk − ωW → 0. Combining these limits, we obtain that lim S(Ak ; wk − ω, Tk ) = 0
(69)
Next, we have |a(Ak ; u0k , ω − wk )| ≤ |a(A; u0k , ω − wk )| + |a(Ak ; u0k , ω − wk ) − a(A; u0k , ω − wk )| = M1 + M2 ; lim M1 = 0, since u0k → u0 in W strongly and ω − wk D 0 weakly in W. In fact, uι0k − uι0 1,Ωι ≤ Cuι0k − uι0 0,∂Ωι → 0
∀ι ≤ s
holds on the basis of the inverse trace theorem. Next, we may write M2 ≤ CB(k) − B∞,Ω u0k W ω − wk W → 0
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since both u0k and wk are bounded in W. Altogether, we arrive at lim a(Ak ; u0k , ω − wk ) = 0 . Making use of Lemma 1, we obtain |jg (Ak ; u0k + ω) − jg (Ak ; u0k + wk )| ≤ C6
(70) ι≤s
ωι − wιk 0,Ωι → 0
(71)
by virtue of the compact trace mapping. From (68)–(71) we obtain lim sup a(Ak ; wk , wk − ω) ≤ 0 and combining this result with (67), we get lim a(Ak ; wk , wk − ω) = 0.
(72)
Then we may write |a(Ak ; wk , w − wk ) − a(A; ω, w − ω)| ≤ |a(Ak ; wk , w − wk ) − a(Ak ; wk , w − ω)| + |a(Ak ; wk , w − ω) − a(A; ω, w − ω)| = I1 + I2 , where I1 = |a(Ak ; wk , ω − wk )| → 0 by (72). Lemma 2 implies I2 → 0. Altogether, we obtain lim a(Ak ; wk , w − wk ) = a(A; ω, w − ω).
(73)
Next, we have |S(Ak ; w − wk , Tk ) − S(A; w − ω, T )| ≤ |S(Ak ; w − wk , Tk ) − S(A; w − wk , T )| + |S(A; w − wk , Tk ) − S(A; w − ω, T )| = J1 + J2 , where J1 ≤ C[F (k) − F 0,Ω + P (k) − P0,Γτ + β(k) − β∞,Ω Tk − T 0,Ω ]w − wk W → 0 since Tk are bounded in W1 by Theorem 1; J2 ≤ |S1 (A; w − wk ) − S1 (A; w − ω)| + |S2 (A; w − wk , Tk − T0 ) − S2 (A; w − ω, T − T0 )| = J21 + J22 , J21 = |S1 (A; ω − wk )| → 0 and J22 → 0 due to the convergence Tk → T in W and as wk D ω weakly in W. Altogether, we obtain lim S(Ak ; w − wk , Tk ) = S(A; w − ω, T ).
(74)
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575
Next, we have |a(Ak ; w − wk , u0k ) − a(A; w − ω, u0 )| ≤ |a(Ak ; w − wk , u0k − u0 )| + |a(Ak ; w − wk , u0 ) − a(A; w − ω, u0 )| = I1 + I2 , I1 ≤ C3 w − wk W u0k − u0 W → 0, using Lemma 1 and the convergence u0k → u0 in W; next, lim I2 = 0 follows from Lemma 2. Thus, we arrive at lim a(Ak ; w − wk , u0k ) = a(A; w − ω, u0 ) .
(75)
Lemma 2 yields that lim[jg (Ak ; u0k + w) − jg (Ak ; u0k + wk )] = jg (A; u0 + w) − jg (A; u0 + ω).
(76)
Let us change the subscript k for n in (64) and pass to the limit with k → ∞. On the basis of (73)–(76), we obtain a(A; ω, w − ω) + jg (A; u0 + w) − jg (A; u0 + ω) ≥ S(A; w − ω, T ) − a(A; w − ω, u0 ).
(77)
Since the variational inequality (17) is uniquely solvable by Proposition 2, ω = w(A) follows from (77). Moreover, the whole sequence {w(An )} tends to w(A) weakly in W. To prove the strong convergence, we write C0 ω − wk 2W ≤ a(An ; ω, ω − wn ) − a(An ; wn , ω − wn ) = K1 + K2 . By Lemma 2, lim K1 = a(A; w, 0) = 0. Since (72) holds for An and wn , lim K2 = lim a(An ; wn , ω − wn ) = 0. As a consequence, w → ω in W, and u(An ) ≡ un = u0n + wn → u0 + ω ≡ u(A) in W, since u0n → u0 in W.
6. Existence of a solution of the worst scenario problem To prove the existence of a solution of the worst scenario problem, we will use the following lemmas.
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Lemma 3. Let Φi (T ), i = 1, 2, be defined by (51) and (52) and let Tn → T in W1 , as n → ∞. Then lim Φi (Tn ) = Φi (T ),
n→∞
i = 1, 2.
Proof. We may write −1
|ϕr (Tn ) − ϕr (T )| ≤ (measN Gr )
Gr
|Tn − T | dx
≤ (measN Gr )−1/2 Tn − T 0,Gr → 0
(78)
and likewise, |ψr (Tn ) − ψr (T )| ≤ (measN−1 Gr )−1/2 Tn − T 0,Gr → 0
(79)
using also the Trace Theorem. Then we have lim Φ1 (Tn ) = lim max ϕr (Tn ) = max lim ϕr (Tn ) = max ϕr (T ) = Φ1 (T ).
n→∞
r≤¯r
r
r
The same argument holds for Φ2 (Tn ), using (79) instead of (78). Lemma 4. Let Φi (u), i = 3, 4, be defined by (53) and (54) and let un → u in W, as n → ∞. Then lim Φi (un ) = Φi (u),
n→∞
i = 3, 4.
Proof. We have −1
|χr (un ) − χr (u)| ≤ measN Gr )
Gr
|uni − ui | |ni (Xr )| dx
≤ (measN Gr )−1/2 un − u0,Gr → 0
(80)
and |χr (un ) − χr (u)| ≤ (measN−1 Gr )−1/2 un − u0,Gr → 0.
(81)
Changing limn→∞ maxr≤¯r to maxr limn as in the proof of Lemma 3, and using (80) or (81), we obtain the assertion of the Lemma. Lemma 5. Let Φi (A; u, T ), i = 5, 6, be defined by (55) and (56). Let An → A in U, An ∈ Uad , un → u in W and Tn → T in L2 (Ω). Then lim Φi (An , un , Tn ) = Φi (A, u, T ),
n→∞
i = 5, 6.
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577
Proof. First, we can show that τij (An ; un , Tn ) − τij (A; u, T )0,Ω → 0
∀i, j ≤ N.
(82)
In fact, we have τij (An ; un , Tn ) − τij (A; u, T )0,Ω ≤ τij (An ; un , Tn ) − τij (An ; u, T )0,Ω + τij (An ; u, T ) − τij (A; u, T )0,Ω = J1 + J2 , J1 ≤ C(un − uW + Tn − T 0,Ω ) → 0, ι since the coefficients cijkl and βijι are bounded in L∞ (Ωι );
J2 ≤ An − AU (uW + T0 − T 0,Ω ) → 0.
Let us consider ωr (A; u, T ) defined in (55) and denote τijn = τij (An ; un , Tn ),
τij = τij (A; u, T ),
so that τijn → τij in L2 (Ω) as n → ∞, follows from (82). Then we have
2 n
I (τ ) − I 2 (τ) dx → 0,
(I 2 (τ n ) − I 2 (τ)) dx ≤ 2 2 2 2
Gr
(83)
Gr
since the integrand is a difference of two quadratic forms in terms of the stress tensor components. In fact, for instance n |τijn τkm − τij τkm | dx Gr
≤ ≤
n (|τijn (τkm − τkm )| + |(τijn − τij )τkm |) dx
Gr n τijn 0 τkm
− τkm 0 + τijn − τij 0 τkm 0 → 0
holds for any two couples (ij) and (km). From (83) we derive that |ωr (An ; un , Tn ) − ωr (A; u, T )| → 0,
∀r ≤ r¯ .
Then we are led to lim Φ5 (An ; un , Tn ) = lim max ωr (An ; un , Tn ) = max lim ωr (An ; un , Tn )
n→∞
n
r
r
n
= max ωr (A; u, T ) = Φ5 (A; u, T ). r
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Next, let µr (A; u, T ) be defined by (56). Then denoting τnn = τijn ni nj , and using (82), we obtain
(−τ n + τn ) dx ≤ n
Gr
Gr
(τij − τ n )ni nj dx ≤ C τij − τijn 0,Ω → 0. ij ij
The rest of the proof for Φ6 is the same as in the previous case. Theorem 3. There exists at least one solution of the worst scenario problems (57) and (58), i = 1, . . . , 6. Proof. Let us denote Ji (A) = Φi (T (A)),
i = 1, 2.
If An ∈ Uad , An → A in U as n → ∞, then An → A in Rp1 and T (An ) → T (A) in W1 by virtue of Theorem 1. Using Lemma 3, we obtain Ji (An ) → Ji (A), so that Ji is continuous on the set Uad . gkl It is easy to show that Uad is compact subset of U, if we employ Arzela–Ascoli Theorem for Uad . As a consequence, Ji attains its maximum on Uad . The same argument can be applied to Ji (A) = Φi (A; u(A), T (A)),
i = 3, . . . , 6.
Here, we employ Theorems 1 and 2 and Lemmas 4 and 5 to verify the continuity of Ji on the set Uad .
7. Conclusion The theory presented in this paper represents extension of results in [8,9] for the case if input data, i.e. thermal conductivity and elastic coefficients, body and surface forces, thermal sources, coefficients of thermal expansion, boundary values, coefficient of friction on contact boundaries, etc. are uncertain. Since the theory extends problems solved in [9], it can be used for mathematical models connected with the safety of construction and of operation of the radioactive waste repositories involving input data which cannot be determined uniquely, but only in some intervals, given by the accuracy of measurements and the approximate solutions of identification problems. The “reliable solution” denotes the worst case among a set of possible solutions where the degree of badness is measured by a criterion functional. For the safety of the high level radioactive waste repositories and other structures under critical conditions we seek the maximal value of this functional, which depends on the solution of the mathematical model. For the computations of such worst scenario problems (maximization of mean values of temperatures, displacements, intensity of shear stresses, principal stresses, stress tensor components, normal and tangential components of the displacement
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579
or stress vector on the contact boundaries, etc.) methods and algorithms known from “optimal design” can be used. For approximations of the problems (P1 ) and (P2 ) we can employ the finite element method (see, e.g. the papers [4,7–9]). For solving the corresponding approximate worst scenario problem the sequential quadratic routine E04UCF from NAG Foundation Library can be recommended. To construct a model of structures under the influence of critical conditions the influence of global tectonics onto a local area, where the critical structure is built, as well as the influence of the resulting local geomechanical processes on a critical structure must be taken into account [9]. Problems of this kind with uncertain input data are problems with high level radioactive waste repositories. In the case of the high level radioactive waste repositories the effects of geodynamical processes in the sense of plate tectonics must be taken into consideration, namely in regions near tectonic areas (e.g. the Japan island arc, the Central and South Europe, etc.), but also in the platform regions (as in Sweden, Canada, etc.). In geomechanics and geodynamics, however, our informations about input data are very questionable as we obtain input data with very small accuracy. Therefore, the presented method as well as algorithms used give a worst scenario (anti-optimal) solution of the problem studied. In practice, it represents a tolerance solution corresponding to the structures in critical situations and, in fact, the obtained solutions facilitate to ensure the high security of constructions and operations of structures under critical situation (e.g. high level radioactive waste repositories, deep mines, etc.). Another example is represented by modelling an interaction between a tunnel wall and a rock massif in the radioactive waste repository tunnels or by modelling of a tunnel crossing by an active deep fault(s), respectively. The theory presented can be used also in mechanics, biomechanics and technology.
Acknowledgements The first author thankfully acknowledges the support of the Grant Agency of the Czech Republic under the Grant 201/01/1200. Both authors acknowledge the support under the Grant COPERNICUSHIPERGEOS II INCO-KIT-977006 and the Grant of the Ministry of Education, Youth and Sports of the Czech Republic no. OK-407.
References [1] Y. Ben-Haim, I.E. Elishakoff, Convex models of uncertainties in applied mechanics, in: Studies in Applied Mechanics, vol. 25, Elsevier, Amsterdam, 1990. [2] J. C´ea, Optimisation, Th´eorie et Algorithmes, Dunod, Paris, 1971. [3] I. Hlav´acˇ ek, Reliable solution of a Signorini contact problem with friction, considering uncertain data, Numer. Linear Algebra Appl. 6 (1999) 411–434. [4] I. Hlav´acˇ ek, J. Nedoma, On a solution of a generalized semi-coercive contact problem in thermo-elasticity, Math. Comput. Simulat. 60 (2002a) 1–17. [5] I. Hlav´acˇ ek, J. Nedoma, Reliable solution of a unilateral frictionless contact problem in quasi-coupled thermo-elasticity with uncertain input data, in: Sloot et al. (Eds.), Proceedings of the Conference ICCS’2002, the Workshop on “Numerical Models in Geomechanics”, Amsterdam, April 2002; Springer Lecture Notes in Computer Sciences, Berlin, Heidelberg, 2002b. [6] J. Neˇcas, Les m´ethodes directes en th´eorie des e´ quations elliptiques, Academia/Masson, Prague/Paris.
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[7] J. Neˇcas, I. Hlav´acˇ ek, Mathematical Theory of Elastic and Elasto-plastic Bodies: An Introduction, Elsevier, Amsterdam, 1981. [8] J. Nedoma, On the Signorini problem with friction in linear thermo-elasticity: the quasi-coupled 2D-case, Appl. Math. 32 3 (1987) 186–199. [9] J. Nedoma, Numerical Modelling in Applied Geodynamics, John Wiley & Sons, New York, 1998. [10] J. Nedoma, J. Haslinger, I. Hlav´acˇ ek, The problem of an obducting lithospheric plate in the Aleutian Arc System: a finite element analysis in the frictionless case, Math. Comput. Model. 12 (1989) 61–75. [11] J. Rohn, Positive definiteness and stability of interval matrices, SIAM J. Matrix Anal. Appl. 15 (1994) 175–184.
Reliable solution of an unilateral contact problem with friction and uncertain data in thermo-elasticity I. Hlav´acˇ ek, J. Nedoma∗ Institute of Computer Science AS CR, Pod Vod´arenskou vˇezˇi´ 2, 182 07 Prague 8, Czech Republic Received 7 July 2004; accepted 2 August 2004 Available online 14 October 2004
Abstract An unilateral contact problem with friction and with uncertain input data in quasi-coupled thermo-elasticity is analysed. As uncertain data coefficients of stress–strain law, coefficients of thermal conductivity, body and surface forces, thermal sources and friction coefficients are assumed, being prescribed in a given set of admissible functions. Method of worst scenario is applied to find the most “dangerous” admissible input data. Stability of the solution with respect to the data is proved and employed for the proof of existence of a solution to the worst scenario problems. Some models in geomechanics, geodynamics and mechanics as well as in technology are stated and the safety of the high level radioactive waste repositories is considered. © 2004 IMACS. Published by Elsevier B.V. All rights reserved. Keywords: Unilateral contact; Steady-state heat flow; Coulomb friction; Finite element analysis; Radioactive waste repositories
1. Introduction In this paper, we will deal with coercive contact problem with friction (see [4,5,7–9]) and uncertain input data in linear quasi-coupled thermo-elasticity, which represents extension of problems solved in [8,9]. Note that a related semi-coercive problem has been treated in a recent paper [4]. By uncertain data we mean input data (physical coefficients, right-hand sides, boundary values, friction limits, etc.), which ∗
Corresponding author. Tel.: +420 266 053 580; fax: +420 286 585 789. E-mail address: [email protected] (J. Nedoma).
0378-4754/$30.00 © 2004 IMACS. Published by Elsevier B.V. All rights reserved. doi:10.1016/j.matcom.2004.08.001
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cannot be determined uniquely but only in some intervals with the errors determined by their measurement errors. The notation reliable solution denotes the worst case among a set of possible solutions, where possibility is given by uncertain input data, and the degree of badness is measured by a criterion-functional (see [3,1]). The main goal of our computation will be to find maximal values of this functional depending on the solution of the problem to be solved. Then we formulate and analyze a corresponding maximization (worst scenario) problem. To its solution we can use methods and algorithms known from optimal design. The solvability of the corresponding worst scenario (maximization) problem is proved. Section 2 is devoted to the formulation of: (i) a stationary heat conduction problem P1 ; and (ii) of the unilateral contact problem with friction P2 , which involves the solution of problem P1 . Sets of admissible uncertain input data are introduced in Section 3 and a unique solvability of problems P1 and P2 is proved for any admissible data. In Section 4, we choose six different criteria, which enable us to find the “worst” data. They represent mean values of: (i) the temperatures or displacements over a priori chosen small subdomains; (ii) the intensity of shear stress; and (iii) the normal component of the stress vector on the contact zone. Section 5 contains the proof of continuous dependence of the temperature and displacement functions on the input data. Existence of at least one solution to each of the worst scenario problems is proved in Section 6.
2. Formulation of the thermo-elastic contact problem Let us consider a union Ω of bounded domains Ωι , ι = 1, . . . , s, with Lipschitz boundaries ∂Ωι , occupied by elastic bodies such that Ω = ∪sι=1 Ωι ⊂ RN , N = {2, 3}. Let the boundary ∂Ω = ∪sι=1 ∂Ωι consist of three disjoint parts Γτ , Γu and Γc , such that ∂Ω = Γ¯ τ ∪ Γ¯ u ∪ Γ¯ c . Assume that (N − 1)-dimensional measures of Γτ , Γu and Γckl are positive, where Γc = ∪k,l Γ kl , Γ kl = ∂Ωk ∩ ∂Ωl , 1 ≤ k, l ≤ s, k = l, and Γ¯ τ , Γ¯ u , Γ¯ c denote the closures in ∂Ω. We will deal with the following quasi-coupled problem of thermo-elasticity, which consists of a pair of boundary value and contact problems to be solved one after the other. 2.1. Problem of stationary heat conduction—problem P1 Let W ι and T1 be given functions. Find the function of temperature T = sι=1 T ι such that ι ∂ ι ∂T κij + W ι = 0 in Ωι , 1 ≤ ι ≤ s, i, j = 1, . . . , N, ∂xi ∂xj κij
∂T ni = 0 on Γu , ∂xj
T = T1 , on Γτ , k l ∂T ∂T k l T =T , κij ni + κij ni = 0 on ∪k,l Γ kl , ∂xj ∂xj
(1) (2) (3)
1 ≤ k, l ≤ s.
(4)
Throughout the paper we use the summation convention, i.e. a repeated index implies summation from 1 to N. Furthermore, nk = (nki ), i = 1, . . . , N, 1 ≤ k ≤ s, denotes the unit normal with respect to
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561
∂Ωk , nk = −nl on Γ kl ; (κijι ) is the matrix of thermal conductivities. Assume that κι are positive definite symmetric matrices, 0 < κ0ι ≤ κijι ζi ζj |ζ|−2
for a.a.
x ∈ Ωι , ζ ∈ RN ,
where κ0ι , are constants independent of x ∈ Ωι . Let κijι ∈ L∞ (Ωι ), W ι ∈ L2 (Ωι ), T1 ∈ sι=1 H 1 (Ωι ), T1k = T1l on ∪k,l Γ kl . Definition 1. We say that a function T is a weak solution of problem P1 , if T − T1 ∈ V1 and b(T, z) = s(z) ∀z ∈ V1 , where b(T, z) =
s ι=1
Ωι
V1 = {z ∈ W1 =
(5)
∂zι dx, ∂xi ∂xj
∂T κijι
ι
sι=1 H 1 (Ωι )
s(z) =
s ι=1
Ωι
W ι zι dx,
| z = 0 on Γτ , zk = zl on ∪k,l Γckl }.
The formulation (5) can be obtained by multiplying Eq. (1) by a test function, integrating by parts over the domain Ωι and using the boundary conditions. 2.2. Problem of unilateral contact problem with friction—problem P2 ι Let the body forces F , the surface tractions P, boundary displacements u0 , elastic coefficients cijkl , ι kl ι coefficients of thermal expansion βij and slip limits gc , the temperature T and the reference temperature T0ι = const. be given. We will deal with the following problem.
Problem 1. (P2 ): Find the displacement field u = (ui ), i = 1, . . . , N, in Ω, such that ∂ τij (uι , T ι ) + Fiι = 0 in Ωι , ∂xj
1 ≤ ι ≤ s,
i = 1, . . . , N,
ι τij (uι , T ι ) = cijkl ekl (uι ) − βijι (T ι − T0ι ) in Ωι , 1 ≤ ι ≤ s,
i = 1, . . . , N,
u = u0 on Γu , i = 1, . . . , N,
τnk ≤ 0,
kl kl |τ kl t | ≤ g on ∪k,l Γ ,
(9)
(ukn − ukn )τnk = 0 on ∪k,l Γ kl ,
1 ≤ k, l ≤ s,
1 ≤ k, l ≤ s,
kl
(12) ukt
ult
=
Here, eij (u) = 1/2((∂ui /∂xj ) + (∂uj /∂xi )), ukn = uki nki , k (uti ), ukti = uki − ukn nki , ult = (ulti ), ulti = uli − uln nki , i = 1, k τnk nki , τnl = τijl nli nkj , τ kl t ≡ τt .
uln
= g =⇒ there exists ϑ ≥ 0 such that
(10) (11)
kl |τ kl =⇒ ukt − ult = 0, t |
|τ kl t |
(7) (8)
τij (u, T )nj = Pi on Γτ , ukn − ukn ≤ 0,
(6)
−
−ϑτ kl t .
(13)
= uli nki = −uli nli (no sum over k or l), ukt . . . , N, τnk = τijk nki nkj , τ kt = (τtik ), τtik = τijk nkj
= −
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ι Assume that cijkl are positive definite symmetric matrices such that ι 0 < c0ι ≤ cijkl ξij ξkl |ξ|−2
x ∈ Ωι , ξ ∈ RN ξij = ξij , 2
for a.a.
ι where c0ι , are constants independent of x ∈ Ωι . Let cijkl ∈ L∞ (Ωι ), Fiι ∈ L2 (Ωι ), Pi ∈ L2 (Γτ ), βijι ∈ ι ∞ ι 1 ι 2 L (Ω ), u0 ∈ [H (Ω )] . To simplify the formulation of stress–strain relations, the entries of any symmetric (N × N) matrix {τij } will be denoted by vector notation {τj }, j = 1, . . . , jN , where jN = N(N + 1)/2, as follows:
τi = τii
for 1 ≤ i ≤ N,
τ3 = τ12
for N = 2,
τ4 = τ23 ,
τ5 = τ31 ,
τ6 = τ12 ,
for N = 3.
Likewise, we replace the symmetric matrices (eij (u)), (βij ) by vectors {ej (u)}, {βj }. Then the stress– strain relation (7) can be rewritten as τi (uι , T ι ) =
jN
Aιij ej (uι ) − βiι (T ι − T0ι ),
1 ≤ i, j ≤ jN , 1 ≤ ι ≤ s,
(14)
j=1
where Aι is a symmetric (jN × jN ) matrix, Aιik ∈ L∞ (Ωι ), ι = 1, . . . , s. Assume that Aιik = 0 for i ≤ N and k > N. It is readily seen that τ : e ≡ τij eij =
N
τi ei + 2
i=1
jN
τi ei .
(15)
i=N+1
Therefore, we can write ι eij ekl cijkl
=
jN
Bijι ei ej ,
i,j=1
where Bι is a symmetric (jN × jN ) matrix such that Bijι = Aιij Bijι
=
2Aιij
Bijι = 0
for 1 ≤ i, j ≤ N,
for N + 1 ≤ i, j ≤ jN .
In what follows, we denote W1 = sι=1 H 1 (Ωι ), sι=1 [H 1 (Ωι )]2 ,
(16)
wW1 =
1/2 wι 21,Ωι
ι≤s
W=
for 1 ≤ i ≤ N, N + 1 ≤ j ≤ jN ,
vW =
ι≤s i≤N
, 1/2
vιi 21,Ωι
.
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563
Assume that the matrices Bι are positive definite, so that 0<
a0ι
≤
jN
Bijι ξi ξj |ξ|−2 ≤ a1ι < +∞
for a.a.
x ∈ Ωι , ξ ∈ RjN ,
i,j=1
where the constants a0ι , a1ι are independent of x ∈ Ωι . Finally, let u0 ∈ W, T0 ∈ W1 , gckl ∈ L∞ (Γ kl ), βjι ∈ L∞ (Ωι ). Let us introduce the space of virtual displacements V = {v ∈ W : v = 0 on Γu } and the set of admissible displacements K = {v ∈ V : vkn − vln ≤ 0 on ∪k,l Γ kl }. Definition 2. We say that the function u is a weak solution of problem (P2 ), if u − u0 ∈ K and a(u, v − u) + jg (v) − jg (u) ≥ S(v − u, T ) where a(u, v) =
s ι=1
jg (v) =
k,l
S(v, T ) =
jN Ωι i,j=1
Γ kl
∀v ∈ u0 + K,
Bijι ei (uι )ej (vι ) dx,
(18)
gkl |vkt − vlt | ds
s ι=1
Ωι
(Fiι vιi
(17)
(19) ι
+ (T −
T0ι )βι
ι
: e(v )) dx +
Γτ
Pi vi ds.
(20)
Remark 1. Note that we have to insert the weak solution T of the problem (P1 ) in S(v, T ). Remark 2. The formulation (17) can be obtained by multiplying Eq. (6) by a test function, integrating by parts over the domain Ωι and using the boundary conditions. In Definition 2 we assume that u0 satisfies conditions uk0n − ul0n = 0 on ∪k,l Γ kl .
(21)
Assumption 1. Throughout the paper, we will assume that measN−1 (Γu ∩ ∂Ωι ) > 0
and
measN−1 (Γτ ∩ ∂Ωι ) > 0 for all ι = 1, . . . , s.
3. Sets of uncertain input data Let us assume that the input data A = {Bι , κι , W ι , T1 , F ι , βι , P, u0 , gkl , ι = 1, . . . , s, ∀k, l}
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are uncertain. Let the only available information about them be that they belong to some sets of admissible data, i.e., ι
B A ∈ Uad ⇔ Bι ∈ Uad , βι
βι ∈ Uad ,
P P ∈ Uad ,
ι
ι
κ κι ∈ Uad ,
W W ι ∈ Uad ,
u0 u0 ∈ Uad ,
gkl ∈ Uad .
T1 T1 ∈ Uad ,
ι
F F ι ∈ Uad ,
gkl
¯ ι exist such that Assume that all the bodies Ωι are piecewise homogeneous, so that partitions of Ω ι ¯ ι = ∪Jj=1 ¯ jι , Ω Ω kl ¯ kl Γ kl = ∪Q q=1 Γq ,
Ωjι ∩ Ωkι = ∅ for j = k, 1 ≤ ι ≤ s ,
(22)
Γqkl ∩ Γpkl = ∅ for q = p, ∀k, l
(23)
and let the data Bι , κι , F ι , W ι , βι be piecewise constant with respect to the partition (22). Let us denote Γu ∩ ∂Ωι = Γuι ,
ι = 1, . . . , s,
(24)
Γτ ∩ ∂Ωι = Γτι ,
ι ≤ s.
(25)
We define the sets of admissible matrices: ι
B ι Uad = {(jN × jN ) symmetric matrices Bι : Bιik (j) ≤ Bik| Ωι
j
= const. ≤
ι ¯ ik B (j),
ι
j≤J,
i, k = 1, . . . , jN },
(26)
¯ ι (j) are given (jN × jN ) symmetric matrices, ι = 1, . . . , s. Assume that positive where Bι (j) and B ι constants cB (j) exist such that ¯ ι (j) − Bι (j)) ≡ cBι (j) for j = 1, . . . , J ι , ι = 1, . . . , s, (27) ¯ ι (j)) − ρ 1 (B λmin 21 (Bι (j) + B 2 where λmin and ρ denotes the minimal eigenvalue and the spectral radius, respectively. ι Remark 3. If the conditions (26) and (27) are fulfilled, the matrices Bι (j) ≡ B|Ω ι are positive definite j
ι
B , ι = 1, . . . , s and any j ≤ J ι . This follows from a result of Rohn (see [11](Theorem for any Bι ∈ Uad 5)).
Likewise, we define the set of admissible matrices ι
κ ι Uad = {(N × N) − symmetric matrices κι : κιik (j) ≤ κik| Ωι
j
= const. ≤
ι κ¯ ik (j),
ι
j ≤ J , i, k ≤ N},
(28)
where κι (j) and κ¯ ι (j) are given (N × N) symmetric matrices, j = 1, . . . , J ι , ι = 1, . . . , s. Assume that positive constants cκι (j) exist such that λmin 21 (κι (j) + κ¯ ι (j)) − ρ 21 (¯κι (j) − κι (j)) ≡ cκι (j) for j ≤ J ι , ι ≤ s , (29) where λmin and ρ denotes the minimal eigenvalue and the spectral radius, respectively. Then the matrices ι κι ι ι κι (j) = κ|Ω ι are positive definite for any κ ∈ U , ι ≤ s, j ≤ J . ad j Next, let us introduce Fι
Uadi = {f ∈ L∞ (Ωι ) : F ιi (j) ≤ f|Ωjι = const. ≤ F¯ iι (j), j ≤ J ι },
(30)
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565
for i ≤ N, ι ≤ s, where F ιi (j) and F¯ iι (j) are given constants; ι
W ¯ ι (j), j ≤ J ι }, = {w ∈ L∞ (Ωι ) : W ι (j) ≤ w|Ωjι = const. ≤ W Uad
(31)
¯ ι (j) are given constants; for ι ≤ s, where W ι (j) and W T1 = {T ∈ L∞ (Γτ ) : T 1 (ι) ≤ T|Γτι = const. ≤ T¯ 1 (ι), ι ≤ s}, Uad
(32)
where T 1 (ι) and T¯ 1 (ι) are given constants; u0i Uad = {u ∈ L∞ (Γu ) : u0i (ι) ≤ u|Γuι = const. ≤ u¯ 0i (ι), ι ≤ s},
(33)
where u0i (ι) and u¯ 0i (ι), i = 1, . . . , N, are given constants. Qιτ ι Let Γτι = ∪q=1 Γ¯ τqι , Γτp ∩ Γτqι = ∅ for p = q. We define Pi Uad = {p ∈ L∞ (Γτ ) : p|Γτqι ∈ C(0),1 (Γ¯ τqι ),
q = 1, . . . , Qιτ , ι = 1, . . . , s, p − pi0 ∞ ≤ C1i , ∂p/∂s∞ ≤ C2i }, where C1i , C2i are given constants and pi0 is a given function such that pi0|Γτqι ∈ C(0),1 (Γ¯ τqι ),
∀q ≤ Qιτ ,
∀ι ≤ s,
∂pi0 /∂s∞ ≤ C2i .
Let ι
β Uadi = {b ∈ L∞ (Ω) : βιi (j) ≤ b|Ωjι = const. ≤ β¯ ι (j), j ≤ J ι },
(34)
for i ≤ jN , ι ≤ s, where βιi (j) and β¯ iι (j) are given constants; gkl
Uad = {g ∈ L∞ (Γ kl ) : g|Γ¯ qkl ∈ C(0),1 (Γ¯ qkl ); 0 ≤ g(s) ≤ g¯ qkl , |dg/ds| ≤ Cqkl a.e. in Γqkl , q ≤ Qkl }, (35) for all pairs kl under consideration, where g¯ qkl and Cqkl are given positive constants. Here, C(0),1 denotes the space of Lipschitz-continuous functions. Finally, we introduce the set of admissible data, as follows: Fι
βι
Pi T1 B κ W Uad = ι≤s Uad × ι≤s Uad × ι≤s,j≤N Uadi × ι≤s Uad × Uad × ι≤s,i≤N Uad × i≤N Uad ι
ι
gkl
u0i × i≤N Uad × k,l Uad .
ι
(36)
T1 Remark 4. To obtain T1 ∈ W1 , we have to extend the boundary values T1 ∈ Uad into the domains k l ι kl Ω properly, i.e. satisfying the conditions T1 = T1 on all Γ . As a consequence at some intersections T1 Γ kl ∩ Γ¯ τ (if any), additional continuity conditions are necessary in the definition of Uad , namely: T1k = T1l , u 0i k l kl if Γ¯ τ ∩ Γ¯ τ = ∅. An analogous remark holds for the data u0i ∈ Uad and Γ ∩ Γ¯ u .
Definition 3. Instead of the bilinear forms and functionals b(T, z), a(u, v), jg (v), s(z), S(v, T ) introduced in Definitions 1 and 2, we will write b(A; T, z), a(A; u, v), jg (A; v), s(A; z), S(A; v, T ) for any A ∈ Uad .
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Lemma 1. There exist positive constants Ci , i = 0, 1, . . . , 6 independent of A ∈ Uad , such that b(A; z, z) ≥ C0 z2W1
∀z ∈ V1 ,
|b(A; z, y)| ≤ C1 zW1 yW1 a(A; v, v) ≥ C2 v2W
∀z, y ∈ W1 ,
∀v ∈ V,
|a(A; v, w)| ≤ C3 vW wW |s(A; z)| ≤ C4 z0,Ω
(37) (38) (39)
∀v, w ∈ W ,
∀z ∈ V1 ,
(40) (41)
|S(A; v, T )| ≤ C5 (v0,Ω + v0,Γτ + T − T0 0,Ω vW ) ∀v ∈ W , uι − vι 0,∂Ωι , ∀u, v ∈ W. |jg (A; u) − jg (A; v)| ≤ C6
(42) (43)
ι≤s
Proof. By Theorem 5 in [11], we have (see (29)) λmin (κι (j)) ≥ cκι (j)
ι
κ ∀κι ∈ Uad ,
ι ≤ s, j ≤ J ι .
As a consequence, we obtain ι b(A; z, z) ≥ min ι cκ (j) |grad zι |2 dx. ι≤s,j≤J
ι≤s
(44)
Ωι
Next, we have |grad zι |2 dx ≥ C1ι zι 21,Ωι
(45)
Ωι
for any restriction zι of z ∈ V1 , making use of Assumption 1. Combining (44) and (45), we arrive at (37). κι The inequality ((38) follows from the definitions of Uad immediately. Arguing as in (44), we may write jN ι a(A; v, v) ≥ min ι cB (j) e2k (vι ) dx. (46) ι≤s,j≤J
ι≤s
Ωι k=1
Using Assumption 1, the Korn’s inequality e(vι ) : e(vι ) dx ≥ C2ι vι 21,Ωι
(47)
Ωι
holds for any restriction vι of v ∈ V . Since jN
1 e2k e:e≤ 2 k=1
(48)
(recall formula (15)), combining (46)–(48), we obtain the inequality (39). The inequality (40) is an easy Bι . consequence of the definitions of Uad
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Likewise, we may write ι ι ¯ |s(A; z)| ≤ maxι {|W (j)|, |W (j)|} |zι | dx ≤ C4 z0,Ω . j≤J
ι≤s
Next, we have |S(A; v, T )| ≤
Ωι
[N 1/2 max{|F¯ iι (j)|, |F ιi (j)|} vι 0,Ωι ι≤s
+ C max β¯ ι (j)
Ωι
ι
2|T −
T0ι | e(vι ) dx
+ C(p0 ) vι 0,Γτι ]
≤ C5 v0,Ω + v0,Γτ + T − T0 0,Ω vW . Finally, we may write |jg (A; u) − jg (A; v)| ≤
Γ kl
k,l
≤
k,l
≤C
gkl |(ukt − vkt ) − (ult − vlt )| ds
max g¯ qkl
q≤Qkl
Γ kl
(|ukt − vkt | + |ult − vlt |) ds
(uki − vki 0,Γ kl + uli − vli 0,Γ kl ) ≤ C6 uι − vι 0,∂Ωι . k,l i≤N
ι≤s
Proposition 1. There exists a unique weak solution T (A) of the problem (P1 ) for any A ∈ Uad . Moreover, T (A)W1 ≤ c,
(49)
where c is independent of A. Proof. We can use the Lax–Milgram Theorem, since (5) can be written as b(t, z) = s(z) − b(T1 , z)
∀z ∈ V1
(50)
for t = T (A) − T1 ∈ V1 . Indeed, the form b is V1 -elliptic due to (37) and continuous due to (38), the right-hand side is a linear continuous functional on V1 , by virtue of (41) and (38). Moreover, inserting z := t tW1 ≤ (C4 + C1 T1 W1 )/C0 follows from (50). Using the inverse theorem on traces (see [6, Theorem 2.5.7]), we obtain ˆ 1ι 1 ,∂Ωι ≤ C ˆC ˜ ι max{|T¯ 1 (ι)|, |T 1 (ι)|} T1ι 1,Ωι ≤ CT 2
for any ι ≤ s. As a consequence, T1 W1 is bounded by a constant independent of A ∈ Uad , so that T (A)W1 ≤ T1 W1 + (C4 + C1 T1 W1 )/C0 ≤ C follows for any A ∈ Uad .
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Proposition 2. There exists a unique weak solution u(A) of problem (P2 ) for any A ∈ Uad . Proof. For brevity, let us omit “A” in what follows. The variational inequality (17) is equivalent with the minimization of the functional Φ(u) = 21 a(u, u) + jg (u) − S(u, T ) over the set u0 + K (cf. [2](Chapter 4, Theorem 0.5)). In fact, u0 + K is a convex set, a(u, u) and S(u, T ) are differentiable with respect to u and jg is convex. By Lemma 1 and the trace theorem, we have |jg (u) − jg (v)| ≤ C6
uι − vι 0,∂Ωι ≤ C
ι≤s
uι − vι 1,Ωι ≤ Cs 2 uι − vι W . 1
ι≤s
As a consequence, the functional Φ is continuous and convex, so that it is weakly lower semi-continuous. The functional is coercive on u0 + K. Indeed, K ⊂ V , so that Lemma 1 yields Φ(u0 + v) = 21 a(u0 + v, u0 + v) − S(u0 + v, T ) ≥ 21 a(v, v) + a(u0 , v) − S(u0 + v, T ) ≥ 21 C2 v2W − C7 vW − C8
∀v ∈ K,
since we may write ˆ |S(u0 + v, T )| ≤ C5 C(u 0 + vW + T − T0 0,Ω u0 + vW ) ˆ ≤ C5 C(u 0 W + vW )(1 + T 0,Ω + T0 0,Ω ) ≤ C7 vW + C8 by virtue of (49). The set u0 + K is weakly closed, being convex and closed in W. As a consequence, there exists a minimizer u(A) of Φ over u0 + K (see [2](Chapter 4, Theorem 0.2)). To verify the uniqueness, assume that u = u0 + v and w = u0 + ω are two solutions of (17). Then we may write w − u = ω − v ∈ V and a(u, w − u) + jg (w) − jg (u) ≥ S(w − u, T ), a(w, u − w) + jg (u) − jg (w) ≥ S(u − w, T ). By addition and Lemma 1, we arrive at C2 w − u2W ≤ a(w − u, w − u) ≤ 0, so that w = u follows.
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4. Criteria of worst scenario To find the “worst”, i.e. the most “dangerous” input data A in the set Uad , we need a criterion, i.e. a functional, which depends on the solution T (A) or u(A) of problem (P1 ) or (P2 ), respectively. Next, we present several examples of such criteria. Let Gr ⊂ ∪ι≤s Ωι , r = 1, . . . , r¯ , be (small) subdomains, adjacent to the boundaries ∂Ωι , for example. We can define Φ1 (T ) = max ϕr (T ),
(51)
r≤¯r
where ϕr (T ) = (measN Gr )−1
Gr
T dx; let Gr ⊂ Γu , r ≤ r¯ and
Φ2 (T ) = max ψr (T ) ,
(52)
r≤¯r
where ψr (T ) = (measN−1 Gr )−1
Gr
T ds. Next, we may define
Φ3 (u) = max χr (u) ,
(53)
r≤¯r
with χr (u) = (measN Gr )−1 Gr ui ni (Xr ) dx, where n(Xr ) is the unit outward normal at a fixed point Xr ∈ ∂Ωι ∩ ∂Gr (if Gr ⊂ Ωι ) to the boundary ∂Ωι ; Φ4 (u) = max χr (u) ;
(54)
r≤¯r
where χr (u) = (measN−1 Gr )−1 G ui ni (Xr ) ds; Gr = ∪ι≤s ∂Ωι \Γu . r Since the weak solution u(A) of our problem (17) depends on T (A), then u(A) = u(A; T (A)) and instead of Φi (u) we write Φi (A; u, T ). Another choice is Φ5 (A; u, T ) = max ωr (A; u, T ) ; r≤¯r
(55)
where ωr (A; u, T ) = (measN Gr )−1 Gr I22 (τ(A; u, T )) dx. Here, I2 (τ) denotes the intensity of shear stress (see, e.g. [7]), i.e. the second fundamental invariant of the stress tensor deviator τ D . We have the formulae I22 (τ) =
3
τijD τijD ,
τijD = τij − 13 τkk δij ;
i,j=1 2 2 2 2 2 2 I22 = 23 [τ11 + τ22 + τ33 − (τ11 τ22 + τ11 τ33 + τ22 τ33 ) + 3(τ12 + τ13 + τ23 )].
In (55), τ(A; u, T ) is defined by the formula (7). For orthotropic material and plane strain, we have to insert τ13 = τ23 = 0. If the friction can be neglected (as in [4,7]), we set g0kl ≡ 0 and define, e.g. Φ6 (A; u, T ) = max µr (A; u, T ); r≤¯r
(56)
where µr (A; u, T ) = (measN Gr )−1 Gr (−τn (A; u, T )) ds, τn = τij ni (Xr )nj (Xr ), Xr ∈ Γc ∩ ∂Gr and Gr is a small subdomain adjacent to Γc (cf. (53)).
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Now we formulate the worst scenario problems: find A0i = arg max Φi (T (A)), A∈Uad
i = 1, 2
(57)
and A0i = arg max Φi (A, u(A), T (A)), A∈Uad
i = 3, . . . , 6,
(58)
where T (A) and u(A) are weak solutions of the problem (P1 ) and (P2 ), respectively.
5. Stability of weak solutions To prove the solvability of worst scenario problems (57) and (58)), we have to study the mapping A → T (A) and A → u(A, T (A)). To these end, we introduce the following decomposition of A ∈ Uad : A = {A , A },
where
A = { ι≤s j≤J ι κι (j), ι≤s j≤J ι W ι (j), ι≤s T1ι },
A ∈ Rp1
p1 = (jN + 1)
J ι + s,
ι≤s
and A = { ι≤s j≤J ι Bι (j), ι≤s j≤J ι F ι (j), ι≤s P ι , ι≤s uι0 , ι≤s j≤J ι βι (j), k,l q≤Qkl gkl (q)}. ι
ι
W κ We are going to show the continuity of the mappings A → T (A ) for A ∈ Uad = ι≤s Uad × ι≤s Uad × T1ι Uad and A → u(A, T (A )) for A ∈ Uad , respectively. , An → A in Rp1 as n → ∞. Then Theorem 1. Let An ∈ Uad
T (An ) → T (A )
in W1 .
Proof. By Proposition 1 the sequence of Tn := T (An ) is bounded. We can choose a weakly convergent subsequence. It is easy to verify that its weak limit coincides with a weak solution T := T (A ). Then the uniqueness implies that Tn D T holds for the whole sequence. To prove its strong convergence, we show that b(A ; Tn − T, Tn − T ) = b(A ; Tn , Tn ) − 2b(A ; Tn , T ) + b(A ; T, T ) → 0
(59)
and make use of the inequality (37). To derive (59), we employ also the “inverse trace theorem” [6](Theorem 2.5.7) in the following form T1 − T1n W1 ≤ CT1 − T1n ∞,Γτ .
(60)
Lemma 2. If An ∈ Uad , An → A in U, where U = Rp1 +p2 × ι≤s q≤Qιτ C(Γ¯ τq ) × k,l q≤Qkl C(Γ¯ qkl ), and un D u weakly in W, then a(An ; un , v) → a(A; u, v)
∀v ∈ W,
(61)
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571
S(An ; un , T ) → S(A; u, T ) ∀T ∈ W1 ,
(62)
jg (An ; un ) → jg (A; u).
(63)
Proof. We have |a(An ; un , v) − a(A; u, v)| ≤ |a(An ; un , v) − a(A; un , v)| + |a(A; un − u, v)| = J1 + J2 ;
jN
ι(n) ι
J1 ≤ (Bik (j) − Bik (j))ei (un )ek (v) dx
ι
ι≤s j≤J ι Ωj i,k=1 ≤ CB(n) − B∞,Ω un W vW → 0, since un are bounded in W; J2 → 0 due to the weak convergence of un . Thus, we are led to (61). Next, we may write |S(An ; un , T ) − S(A; u, T )| ≤ |S(An ; un , T ) − S(A; un , T )| + |S(An ; un − u, T )| = I1 + I2 ;
ι(n) ι(n) ι ι ι ι ι ι(n) ι I1 ≤
(Fi (j) − Fi (j))uni + (T − T0 )(β − β ) : e(un )) dx + (Pi −Pi )uni ds
Ωjι
Γτ ι ι≤s,j≤J
≤ C(F (n) − F ∞,Ω + T − T0 W1 β(n) − β∞,Ω + P (n) − P∞,Γτ )un W → 0, I2 → 0 by virtue of the weak convergence of un . As a consequence, (62) follows. We have |jg (An ; un ) − jg (A; u)| ≤ |jg (An ; un ) − jg (A; un )| + |jg (A; un ) − jg (A; u)| = K1 + K2 , K1 ≤
k,l
Γ kl
kl
(g − gkl ) uk − ul ds ≤ C gnkl − gkl ∞,Γ kl un W → 0. n nt nt k,l
Finally, (43) and the compactness of the trace map H 1 (Ωι ) → L2 (∂Ωι ) yield that K2 ≤ C6
ι≤s
uι − uιn 0,∂Ωι → 0.
Thus, we obtain (63). Theorem 2. Then
Let An ∈ Uad , An → A in U ≡ Rp1 +p2 × ι≤s q≤Qιτ C(Γ¯ τq ) × k,l q≤Qkl C(Γ¯ qkl ).
u(An ) → u(A) in W.
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Proof. For brevity, denote un := u(An ), u := u(A), u0n := u0 (An ), u0 := u0 (A), Tn := T (An ), T := T (A), in what follows. We shall insert u := u0 + w(A),
w(A) ∈ K,
un := u0n + wn
wn ∈ K
v := u0 + w or v := u0n + w, w ∈ K into the variational inequality (17). Thus, we obtain a(An ; wn , w − wn ) + jg (An ; u0n + w) − jg (An ; u0n + wn ) ≥ S(An ; w − wn , Tn ) − a(An ; u0n , w − wn ).
(64)
Taking w = 0 and using Lemma 1, we get C0 wn 2W ≤ a(An ; wn ; wn ) + jg (An ; uon + wn ) ≤ S(An ; wn , Tn ) + a(An ; u0n , −wn ) + jg (An ; u0n ) ≤ C5 (wn 0,Ω + wn 0,Γτ + Tn − T0 0,Ω wn W ) uιon 0,∂Ωι . + C3 u0n W wn W + C6
(65)
ι≤s
u0i , and Remark 4, the extensions uιon are By Theorem 1, Tn are bounded in W1 . By definition of Uad 1 ι bounded in H (Ω ). Then
C0 wn 2W ≤ C7 wn W + C8 follows from (65). As a consequence, wn are bounded in W and there exists a subsequence {wk } and a function ω ∈ W such that wk D ω weakly in W, as k → ∞.
(66)
We can show that ω = w(A). First, ω ∈ K since the set K is weakly closed. Second, we have a(Ak ; wk − ω, wk − ω) ≥ 0 so that a(Ak ; wk , wk − ω) ≥ a(Ak ; ω, wk − ω). By Lemma 2, we obtain lim inf a(Ak ; wk , wk − ω) ≥ lim a(Ak ; ω, wk − ω) = 0 . Inserting w := ω into (64), we get a(Ak ; wk , ω − wk ) + jg (Ak ; uok + ω) − jg (Ak ; uok + wk ) ≥ S(Ak ; ω − wk , Tk ) − a(Ak ; u0k , ω − wk )
(67)
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and lim sup a(Ak ; wk , wk − ω) ≤ lim sup S(Ak ; wk − ω, Tk ) + lim sup a(Ak ; u0k , ω − wk ) + lim sup[jg (Ak ; uok + ω) − jg (Ak ; uok + wk )].
(68)
For any A ∈ Uad , v ∈ W and T ∈ W1 , we may write S(A; v, T ) = S1 (A; v) + S2 (A; T − T0 , v), where S1 (A; ·) is a linear continuous functional on W and S2 (A; ·, ·) is a bilinear continuous form on L2 (Ω) × W. Then we have S(Ak ; wk − ω, Tk ) = S1 (Ak ; wk − ω) + S2 (Ak ; Tk − T0 , wk − ω) = S1 (A; wk − ω) + [S1 (Ak ; wk − ω) − S1 (A; wk − ω)] + S2 (A; Tk − T0 , wk − ω) + [S2 (Ak ; Tk − T0 , wk − ω) − S2 (A; Tk − T0 , wk − ω)]. By virtue of the weak convergence, S1 (A; wk − ω) → 0;
S2 (A; Tk − T0 , wk − ω) → S2 (A; T − T0 , 0) = 0
since Tk → T in W1 by virtue of Theorem 1. Next, we have |S1 (Ak ; wk − ω) − S1 (A; wk − ω)| ≤ CAk − AU wk − ωW → 0, |S2 (Ak ; Tk − T0 , wk − ω) − S2 (A; Tk − T0 , wk − ω)| ≤ CAk − AU Tk − T 0,Ω wk − ωW → 0. Combining these limits, we obtain that lim S(Ak ; wk − ω, Tk ) = 0
(69)
Next, we have |a(Ak ; u0k , ω − wk )| ≤ |a(A; u0k , ω − wk )| + |a(Ak ; u0k , ω − wk ) − a(A; u0k , ω − wk )| = M1 + M2 ; lim M1 = 0, since u0k → u0 in W strongly and ω − wk D 0 weakly in W. In fact, uι0k − uι0 1,Ωι ≤ Cuι0k − uι0 0,∂Ωι → 0
∀ι ≤ s
holds on the basis of the inverse trace theorem. Next, we may write M2 ≤ CB(k) − B∞,Ω u0k W ω − wk W → 0
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since both u0k and wk are bounded in W. Altogether, we arrive at lim a(Ak ; u0k , ω − wk ) = 0 . Making use of Lemma 1, we obtain |jg (Ak ; u0k + ω) − jg (Ak ; u0k + wk )| ≤ C6
(70) ι≤s
ωι − wιk 0,Ωι → 0
(71)
by virtue of the compact trace mapping. From (68)–(71) we obtain lim sup a(Ak ; wk , wk − ω) ≤ 0 and combining this result with (67), we get lim a(Ak ; wk , wk − ω) = 0.
(72)
Then we may write |a(Ak ; wk , w − wk ) − a(A; ω, w − ω)| ≤ |a(Ak ; wk , w − wk ) − a(Ak ; wk , w − ω)| + |a(Ak ; wk , w − ω) − a(A; ω, w − ω)| = I1 + I2 , where I1 = |a(Ak ; wk , ω − wk )| → 0 by (72). Lemma 2 implies I2 → 0. Altogether, we obtain lim a(Ak ; wk , w − wk ) = a(A; ω, w − ω).
(73)
Next, we have |S(Ak ; w − wk , Tk ) − S(A; w − ω, T )| ≤ |S(Ak ; w − wk , Tk ) − S(A; w − wk , T )| + |S(A; w − wk , Tk ) − S(A; w − ω, T )| = J1 + J2 , where J1 ≤ C[F (k) − F 0,Ω + P (k) − P0,Γτ + β(k) − β∞,Ω Tk − T 0,Ω ]w − wk W → 0 since Tk are bounded in W1 by Theorem 1; J2 ≤ |S1 (A; w − wk ) − S1 (A; w − ω)| + |S2 (A; w − wk , Tk − T0 ) − S2 (A; w − ω, T − T0 )| = J21 + J22 , J21 = |S1 (A; ω − wk )| → 0 and J22 → 0 due to the convergence Tk → T in W and as wk D ω weakly in W. Altogether, we obtain lim S(Ak ; w − wk , Tk ) = S(A; w − ω, T ).
(74)
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Next, we have |a(Ak ; w − wk , u0k ) − a(A; w − ω, u0 )| ≤ |a(Ak ; w − wk , u0k − u0 )| + |a(Ak ; w − wk , u0 ) − a(A; w − ω, u0 )| = I1 + I2 , I1 ≤ C3 w − wk W u0k − u0 W → 0, using Lemma 1 and the convergence u0k → u0 in W; next, lim I2 = 0 follows from Lemma 2. Thus, we arrive at lim a(Ak ; w − wk , u0k ) = a(A; w − ω, u0 ) .
(75)
Lemma 2 yields that lim[jg (Ak ; u0k + w) − jg (Ak ; u0k + wk )] = jg (A; u0 + w) − jg (A; u0 + ω).
(76)
Let us change the subscript k for n in (64) and pass to the limit with k → ∞. On the basis of (73)–(76), we obtain a(A; ω, w − ω) + jg (A; u0 + w) − jg (A; u0 + ω) ≥ S(A; w − ω, T ) − a(A; w − ω, u0 ).
(77)
Since the variational inequality (17) is uniquely solvable by Proposition 2, ω = w(A) follows from (77). Moreover, the whole sequence {w(An )} tends to w(A) weakly in W. To prove the strong convergence, we write C0 ω − wk 2W ≤ a(An ; ω, ω − wn ) − a(An ; wn , ω − wn ) = K1 + K2 . By Lemma 2, lim K1 = a(A; w, 0) = 0. Since (72) holds for An and wn , lim K2 = lim a(An ; wn , ω − wn ) = 0. As a consequence, w → ω in W, and u(An ) ≡ un = u0n + wn → u0 + ω ≡ u(A) in W, since u0n → u0 in W.
6. Existence of a solution of the worst scenario problem To prove the existence of a solution of the worst scenario problem, we will use the following lemmas.
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Lemma 3. Let Φi (T ), i = 1, 2, be defined by (51) and (52) and let Tn → T in W1 , as n → ∞. Then lim Φi (Tn ) = Φi (T ),
n→∞
i = 1, 2.
Proof. We may write −1
|ϕr (Tn ) − ϕr (T )| ≤ (measN Gr )
Gr
|Tn − T | dx
≤ (measN Gr )−1/2 Tn − T 0,Gr → 0
(78)
and likewise, |ψr (Tn ) − ψr (T )| ≤ (measN−1 Gr )−1/2 Tn − T 0,Gr → 0
(79)
using also the Trace Theorem. Then we have lim Φ1 (Tn ) = lim max ϕr (Tn ) = max lim ϕr (Tn ) = max ϕr (T ) = Φ1 (T ).
n→∞
r≤¯r
r
r
The same argument holds for Φ2 (Tn ), using (79) instead of (78). Lemma 4. Let Φi (u), i = 3, 4, be defined by (53) and (54) and let un → u in W, as n → ∞. Then lim Φi (un ) = Φi (u),
n→∞
i = 3, 4.
Proof. We have −1
|χr (un ) − χr (u)| ≤ measN Gr )
Gr
|uni − ui | |ni (Xr )| dx
≤ (measN Gr )−1/2 un − u0,Gr → 0
(80)
and |χr (un ) − χr (u)| ≤ (measN−1 Gr )−1/2 un − u0,Gr → 0.
(81)
Changing limn→∞ maxr≤¯r to maxr limn as in the proof of Lemma 3, and using (80) or (81), we obtain the assertion of the Lemma. Lemma 5. Let Φi (A; u, T ), i = 5, 6, be defined by (55) and (56). Let An → A in U, An ∈ Uad , un → u in W and Tn → T in L2 (Ω). Then lim Φi (An , un , Tn ) = Φi (A, u, T ),
n→∞
i = 5, 6.
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577
Proof. First, we can show that τij (An ; un , Tn ) − τij (A; u, T )0,Ω → 0
∀i, j ≤ N.
(82)
In fact, we have τij (An ; un , Tn ) − τij (A; u, T )0,Ω ≤ τij (An ; un , Tn ) − τij (An ; u, T )0,Ω + τij (An ; u, T ) − τij (A; u, T )0,Ω = J1 + J2 , J1 ≤ C(un − uW + Tn − T 0,Ω ) → 0, ι since the coefficients cijkl and βijι are bounded in L∞ (Ωι );
J2 ≤ An − AU (uW + T0 − T 0,Ω ) → 0.
Let us consider ωr (A; u, T ) defined in (55) and denote τijn = τij (An ; un , Tn ),
τij = τij (A; u, T ),
so that τijn → τij in L2 (Ω) as n → ∞, follows from (82). Then we have
2 n
I (τ ) − I 2 (τ) dx → 0,
(I 2 (τ n ) − I 2 (τ)) dx ≤ 2 2 2 2
Gr
(83)
Gr
since the integrand is a difference of two quadratic forms in terms of the stress tensor components. In fact, for instance n |τijn τkm − τij τkm | dx Gr
≤ ≤
n (|τijn (τkm − τkm )| + |(τijn − τij )τkm |) dx
Gr n τijn 0 τkm
− τkm 0 + τijn − τij 0 τkm 0 → 0
holds for any two couples (ij) and (km). From (83) we derive that |ωr (An ; un , Tn ) − ωr (A; u, T )| → 0,
∀r ≤ r¯ .
Then we are led to lim Φ5 (An ; un , Tn ) = lim max ωr (An ; un , Tn ) = max lim ωr (An ; un , Tn )
n→∞
n
r
r
n
= max ωr (A; u, T ) = Φ5 (A; u, T ). r
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Next, let µr (A; u, T ) be defined by (56). Then denoting τnn = τijn ni nj , and using (82), we obtain
(−τ n + τn ) dx ≤ n
Gr
Gr
(τij − τ n )ni nj dx ≤ C τij − τijn 0,Ω → 0. ij ij
The rest of the proof for Φ6 is the same as in the previous case. Theorem 3. There exists at least one solution of the worst scenario problems (57) and (58), i = 1, . . . , 6. Proof. Let us denote Ji (A) = Φi (T (A)),
i = 1, 2.
If An ∈ Uad , An → A in U as n → ∞, then An → A in Rp1 and T (An ) → T (A) in W1 by virtue of Theorem 1. Using Lemma 3, we obtain Ji (An ) → Ji (A), so that Ji is continuous on the set Uad . gkl It is easy to show that Uad is compact subset of U, if we employ Arzela–Ascoli Theorem for Uad . As a consequence, Ji attains its maximum on Uad . The same argument can be applied to Ji (A) = Φi (A; u(A), T (A)),
i = 3, . . . , 6.
Here, we employ Theorems 1 and 2 and Lemmas 4 and 5 to verify the continuity of Ji on the set Uad .
7. Conclusion The theory presented in this paper represents extension of results in [8,9] for the case if input data, i.e. thermal conductivity and elastic coefficients, body and surface forces, thermal sources, coefficients of thermal expansion, boundary values, coefficient of friction on contact boundaries, etc. are uncertain. Since the theory extends problems solved in [9], it can be used for mathematical models connected with the safety of construction and of operation of the radioactive waste repositories involving input data which cannot be determined uniquely, but only in some intervals, given by the accuracy of measurements and the approximate solutions of identification problems. The “reliable solution” denotes the worst case among a set of possible solutions where the degree of badness is measured by a criterion functional. For the safety of the high level radioactive waste repositories and other structures under critical conditions we seek the maximal value of this functional, which depends on the solution of the mathematical model. For the computations of such worst scenario problems (maximization of mean values of temperatures, displacements, intensity of shear stresses, principal stresses, stress tensor components, normal and tangential components of the displacement
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or stress vector on the contact boundaries, etc.) methods and algorithms known from “optimal design” can be used. For approximations of the problems (P1 ) and (P2 ) we can employ the finite element method (see, e.g. the papers [4,7–9]). For solving the corresponding approximate worst scenario problem the sequential quadratic routine E04UCF from NAG Foundation Library can be recommended. To construct a model of structures under the influence of critical conditions the influence of global tectonics onto a local area, where the critical structure is built, as well as the influence of the resulting local geomechanical processes on a critical structure must be taken into account [9]. Problems of this kind with uncertain input data are problems with high level radioactive waste repositories. In the case of the high level radioactive waste repositories the effects of geodynamical processes in the sense of plate tectonics must be taken into consideration, namely in regions near tectonic areas (e.g. the Japan island arc, the Central and South Europe, etc.), but also in the platform regions (as in Sweden, Canada, etc.). In geomechanics and geodynamics, however, our informations about input data are very questionable as we obtain input data with very small accuracy. Therefore, the presented method as well as algorithms used give a worst scenario (anti-optimal) solution of the problem studied. In practice, it represents a tolerance solution corresponding to the structures in critical situations and, in fact, the obtained solutions facilitate to ensure the high security of constructions and operations of structures under critical situation (e.g. high level radioactive waste repositories, deep mines, etc.). Another example is represented by modelling an interaction between a tunnel wall and a rock massif in the radioactive waste repository tunnels or by modelling of a tunnel crossing by an active deep fault(s), respectively. The theory presented can be used also in mechanics, biomechanics and technology.
Acknowledgements The first author thankfully acknowledges the support of the Grant Agency of the Czech Republic under the Grant 201/01/1200. Both authors acknowledge the support under the Grant COPERNICUSHIPERGEOS II INCO-KIT-977006 and the Grant of the Ministry of Education, Youth and Sports of the Czech Republic no. OK-407.
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